As part of a wedding photoshoot, a photographer places one end of a 6.00 m ladde
ID: 1418907 • Letter: A
Question
As part of a wedding photoshoot, a photographer places one end of a 6.00 m ladder of mass 8.50 kg against a house and the other end on a concrete pad with the base of the ladder 4.00 m from the house (assume the wall the ladder rests against is frictionless and the concrete pad has a coefficient of static friction s with the base of the ladder). The ladder is uniform, so its center of mass is at its geometric center. If the 69.4 kg groom stands on the ladder 2.00 m from the bottom and the 58.6 kg bride stands 4.00 m from the bottom, what normal force must be exerted by the wall on the top of the ladder? What is the minimum coefficient of static friction s needed to keep the ladder from slipping?
Explanation / Answer
use the point the ladder contact with the floor as the pivotal point, the torque equation requires
the angle between the ladder and the floor is 48.19 degrees
N*6*sin48.19 = (8.5 *3 +69.4*2.0+58.6*4 )9.8*cos48.19
N = 582.45 N
the normal force of the floor is
N' = (8.5+69.4+58.6 )9.8 = 1337.7 N
the maximum static friction is
us*N' >= N = 582.45
us>= 582.45/1337.7 = 0.435
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